\newproblem{lay:1_1_14}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.1.14}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Clara Susana Rey Abad, Nov. 4, 2013} \\}{}

  % Problem statement
	
	Solve the equation system:
	\begin{center}
	 $\begin{array}{rcr}
	  2x_1      -6x_3&=&-8\\
		       x_2+2x_3&=& 3\\
		3x_1+ 6x_2-2x_3&=&-4\\
	  \end{array}$
	\end{center}
	
}{
   % Solution
	Let us construct the augmented system matrix
	\begin{center}
		$\left(\begin{array}{rrr|r}
		   2 & 0 & -6 & -8 \\
			 0 & 1 &  2 &  3 \\
			 3 & 6 & -2 & -4
		\end{array}\right)$
	\end{center}
	
	Now, we apply row operations to solve it
	\begin{center}
		\begin{tabular}{cc}
			 $\mathbf{r}_1\leftarrow \mathbf{r}_1\div 2$ &
			 $\left(\begin{array}{rrr|r}
				 1 & 0 & -3 & -4\\
				 0 & 1 &  2 &  3 \\
				 3 & 6 & -2 & -4		
				 \end{array}\right)$ \\
			 $\mathbf{r}_3\leftarrow \mathbf{r}_3-3\mathbf{r}_1$ &
			 $\left(\begin{array}{rrr|r}
				 1 &  0  &  -3 & -4 \\
				 0 &  1  &   2 &  3 \\
				 0 &  6  &   4 &  8
			 \end{array}\right)$ \\
			
			$\mathbf{r}_3\leftarrow \mathbf {r}_3\div 2$ &
			$\left(\begin{array}{rrr|r}
			   1 &  0 &  -3  & -4 \\
				 0 &  1 &   2  &  3 \\
				 0 &  3 &   2  &  4
				\end{array}\right)$\\
				
			$\mathbf{r}_3\leftarrow \mathbf {r}_3-3\mathbf{r}_2$ &
			$\left(\begin{array}{rrr|r}
			   1 &  0 &  -3  & -4 \\
				 0 &  1 &   2  &  3 \\
				 0 &  0 &  -4  & -5
				\end{array}\right)$ \\
				
			$\mathbf{r}_3\leftarrow \mathbf {r}_3\div 4$ &
			$\left(\begin{array}{rrr|r}
			   1 &  0 & -3 & -4 \\
				 0 &  1 &  2 &  3  \\
				 0 &  0 & -1 & -5/4 
				\end{array}\right)$ \\
			$\mathbf{r}_2\leftarrow \mathbf {r}_2+2\mathbf{r}_3$ &
			$\left(\begin{array}{rrr|r}
			   1 & 0 & -3 & -4 \\
				 0 & 1 &  0 & 1/2 \\
				 0 & 0 & -1 &-5/4
				\end{array}\right)$\\
			$\mathbf{r}_1\leftarrow \mathbf {r}_1-3\mathbf{r}_3$ &
			$\left(\begin{array}{rrr|r}
			   1 & 0 &  0 & -1/4\\
				 0 & 1 &  0 &  1/2\\
				 0 & 0 & -1 & -5/4
				\end{array}\right)$\\
			$\mathbf{r}_3\leftarrow \mathbf {r}_3\div -1$ &
			$\left(\begin{array}{rrr|r}
			   1 & 0 &  0 & -1/4\\
				 0 & 1 &  0 &  1/2\\
				 0 & 0 &  1 &  5/4
				\end{array}\right)$\\
		\end{tabular}
	\end{center}
	
	Whe can deduce from the reduced echelon form that
	\begin{center}
	 $\begin{array}{rcr}
	  x_1 &=&-1/4\\
		x_2 &=& 1/2\\
		x_3 &=& 5/4\\
	  \end{array}$
	\end{center}
	
	 Therefore, there is a unique solution of the system. The equation system is compatible determinated.

}
\useproblem{lay:1_1_14}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
